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zzr0ck3r

  • one year ago

Good methods for finding eigen values? \(M = \begin{array}{cc} 0&1&1&1&1& \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1\\ 1&1&1&1&0 \end{array}\)

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  1. zzr0ck3r
    • one year ago
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    @freckles where you go ;(

  2. UsukiDoll
    • one year ago
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    \[\det(A- \lambda I) =0\] and then use cofactor expansion like crazy

  3. zzr0ck3r
    • one year ago
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    I said good methods :)

  4. UsukiDoll
    • one year ago
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    =(

  5. UsukiDoll
    • one year ago
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    row operations first? idk that may destroy the determinant...

  6. freckles
    • one year ago
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    lol well ... I was kind of trying to see if could find a pattern but I didn't see a pattern like you know M_n is a n by n matrix with 0's on the diagonal and 1's everywhere else but it was totally not a good method

  7. freckles
    • one year ago
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    and not just a bad method but it didn't seem to work like that

  8. UsukiDoll
    • one year ago
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    does this whole matrix break one of the rules for row operations? Don't we need a 1 at the top left corner of the first row?

  9. zzr0ck3r
    • one year ago
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    word

  10. UsukiDoll
    • one year ago
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    row swaps... so it does not break the first rule

  11. UsukiDoll
    • one year ago
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    I would switch the last one with the first one r_1<->r_5

  12. UsukiDoll
    • one year ago
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    swap r_2 <-> r_4 , then r_3 <-> r_4 ?

  13. freckles
    • one year ago
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    actually I do see a pattern

  14. freckles
    • one year ago
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    so there must be some kinda of shortcut for M_n where M_n is a n by n matrix with 0's a long the diagonal and 1's elsewhere

  15. freckles
    • one year ago
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    I want to make a conjecture \[\text{ egienvalues }(M_n)=n-1,-1 \text{ (multiplicity } (n-1) \text{ )}\]

  16. freckles
    • one year ago
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    Now I know this isn't a method but it might help to come up with something from this if it is even true

  17. freckles
    • one year ago
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    like we know the trace of M is suppose to be the sum of the eigenvalues

  18. freckles
    • one year ago
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    \[0=\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5\]

  19. freckles
    • one year ago
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    I don't know a quick way to find the determinant but the product of the eigenvalues equals the determinant

  20. zzr0ck3r
    • one year ago
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    ill play with this too

  21. freckles
    • one year ago
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    another cute thing \[M_n=[1]_n-I\] where 1_n is a n by n matrix only made of 1's and I is the identity matrix don't know if this helps yet

  22. freckles
    • one year ago
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    you know I'm not totally sure yet but it seems to be: in this case anyways \[\text{ eigenvalues(}M_n\text{)}=\text{ egivenvalues(}[1]_n\text{)}-\text{ egienvalues(}I\text{)}\]

  23. freckles
    • one year ago
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    the eigenvalues of I are 1 (n amount of times) and the eigenvalues of [1]_n are n, 0 (n-1 amount of times)

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